1. Thermochemistry

2.  What is energy?  Energy is the ability to do work or produce heat.  The Law of Conservation of Energy: ◦ This law states that can not be created or destroyed only transferred.  Two types of energy: ◦ Kinetic ◦ Potential

3.  Heat is energy transferred from a warmer object to a cooler object.  Heat is represented mathematically as q  The amount of heat energy required to raise the temperature of one gram of pure water by one degree Celsius is called…  a calorie  The SI unit for energy is the Joule  1 Joule = 0.2390 calories  1 calorie = 4.184 Joules

4.  A breakfast of cereal, orange juice, and milk contains 230 Calories. Convert this amount of energy in to Joules.  9.6 x 10 5 J

5.  Glucose is a simple sugar found in fruit. Burning 1.00 g of glucose releases 15.6 kJ of energy. How many Calories are released?  3.73 Calories  An fruit and oatmeal bar contains 142 Calories. Convert this energy to Joules  5.94 x 10 5  A chemical reaction releases 86.5 kJ of heat. How many Calories are released?  20.7 Calories

6.  The specific heat of any substance is the amount of heat required to raise the temperature of one gram of that substance by 1 degree Celsius.  The specific heat of water is:  1 cal/g- o C  4.184 J/g- o C  The specific heat of concrete is 0.84 J/g- o C  This is why in the summer concrete gets hot and water says cool.

7.  The specific heat of a substance can be used to calculate the heat energy absorbed or given off when that substance changes temperature.  q = (C)(m)(ΔT)  remember ΔT = T final – T initial  Calculate the heat absorbed by a 5 x 10 3 g block of concrete when its temperature is raised from 20 o C to 26 o C.  25,000 J or 25 kJ

8.  The temperature of a sample of iron with a mass of 10.0 g changed from 50.4 o C to 25.0 o C and released 114 J. What is the specific heat of iron?  0.449 J/g- o C  If the temperature of 34.4 g of ethanol increase from 25.0 o C to 78.8 o C, how much heat has been absorbed buy the ethenol. (C for ethanol is 2.44 J/g- o C)  4.52 x 10 3 J

9.  A 155 g sample of an unknown substance was heated from 25.0 o C to 40.0 o C. In the process, the substance absorbed 5696 J of energy. What is the specific heat of the substance?  2.45 J/g- o C  A 38.8 g piece of metal alloy absorbs 181 J as its temperature increases from 25 o C to 36 o C. What is the alloy’s specific heat?  0.424 J/g- o C

10.  Thermochemistry is the study of heat changes during chemical reactions or phase changes.  When studying thermochemistry we look at two things:  System ◦ The system is the specific part of the universe that we are studying.  Surroundings ◦ The surroundings are everything else in the universe.

11.  Enthalpy is defined as the heat content of a system at constant pressure.  The change in enthalpy for a reaction is called the enthalpy (heat) of reaction.  ΔH rxn  ΔH rxn = H products – H reactnats

12.  A thermochemical equation is a balanced equation that includes the physical states of all reactants and products and the enthalpy change.  4 Fe(s) + 3 O 2 (g)  2 Fe 2 O 3 (s)ΔH = -1625 kJ  NH 4 NO 3 (s)  NH 4 + (aq) + NO 3 - ΔH = 27 kJ

13.  Hess’s Law states that if you can add two or more equations to produce a final equation for a reaction than the sum of the enthalpy changes of the individual reactions is the enthalpy change of the overall reaction.

14.  Calculate ΔH for the reaction  2 H 2 O 2 (l)  2 H 2 O(l) + O 2 (g)  2 H 2 (g) + O 2 (g)  2 H 2 O(l)ΔH = -572 kJ  H 2 (g) + O 2 (g)  H 2 O 2 (l)ΔH = -188 kJ

15.  Use equations (a) and (b) to determine ΔH for the following reaction:  2 CO(g) + 2 NO(g)  2 CO 2 (g) + N 2 (g) a) 2 CO(g) + O 2 (g)  2 CO 2 (g) ΔH = -566.0 kJ a) N 2 (g) + O 2 (g)  2NO(g) ΔH = -180.6 kJ

16.  ΔH for the following reaction is -1789 kJ. Use equation (a) to determine ΔH for reaction (b).  4 Al(s) + 3 MnO 2 (s)  2 Al 2 O 3 (s) + 3 Mn(s) a) 4 Al(s) + 3 O 2  2 Al 2 O 3 (s)ΔH = -3352 kJ b) Mn(s) + O 2 (g)  MnO 2 ΔH = ?

17.  The enthalpy of formation for any reaction is defined as the heat change when all reactants are in their elemental form and only one mole of product is produced.  S(s) + 3 F 2 (g)  SF 6 ΔH o f = -1220 kJ  Sometimes we need to use fractional coefficients.  H 2 (g) + F 2 (g)  HFΔH o f = -273 kJ

18.  We can use the enthalpy of formation for components of a reaction to calculate the total enthalpy change of the reaction (ΔH rxn ).  H 2 S(g) + 4 F 2 (g)  2 HF(g) + SF 6 (g) a) ½ H 2 (g) + ½ F 2 (g)  HF ΔH o f = -273 kJ b) S(s) + 3 F 2 (g)  SF 6 ΔH o f = -1220 kJ c) H 2 (g) + S(s)  H 2 S(g) ΔH o f = -21 kJ

19.  Determine ΔH for CH 4 (g) + 2 O 2 (g)  CO 2 (g) + 2 H 2 O(l)  Using:  ΔH o f (CO 2 ) = -394 kJ  ΔH o f (H 2 O) = -286 kJ  ΔH o f (CH 4 ) = -75 kJ  ΔH o f (O 2 ) = 0 kJ

20.  When things rust the reaction taking place is:  4 Fe(s) + 3 O 2 (g)  2 Fe 2 O 3 (s)ΔH = -1625 kJ  Any physical or chemical that occurs with no outside intervention is a spontaneous process.  If we reverse the above reaction:  2 Fe 2 O 3 (s)  3 O 2 (g) + 4 Fe(s) ΔH = 1625 kJ  This process would be non-spontaneous

21.  Entropy is a measure of the number of possible ways a system can be configured.  If we have a piece of paper cut into 8 different sections there would be 56 different ways we could arrange them. (8 x 7)  If we cut the paper into 16 different pieces there would be 240 different ways we could arrange them. (16 x 15)  We have increased the papers entropy.

22.  The second law of thermodynamics states that a spontaneous reaction will always occur in such a way that entropy increases.  Remember that the change in enthalpy (ΔH) is defined as:  H products – H reactants  Similarly:  ΔS = S products – S reactants  If ΔS is positive the entropy of the system is increasing.  If ΔS is negative the entropy of the system is decreasing.

23. Phase Changes: ◦ When a phase change occurs from a more ordered state to a less ordered state ΔS will be positive. ◦ Solid  Liquid ΔS > 0 ◦ When a phase change occurs from a less ordered state to a more ordered state ΔS will be negative. ◦ Gas  LiquidΔS < 0 Dissolving a gas into a solvent always results in a decrease in entropy.

24.  Assuming no change in physical state, entropy increases when the number of moles of products is greater than the number of moles of reactants. ◦ 2 SO 3 (g)  2 SO 2 (g) + O 2 (g)  Entropy increases when a solute dissolves in a solvent. ◦ NaCl(s)  Na + (aq) + Cl - (aq)  Entropy increases as temperature increases.

25.  Predict the sign of ΔS for each of the following chemical of physical processes  ClF(g) + F 2 (g)  ClF 3 (g)ΔS =  NH 3 (g)  NH 3 (aq)ΔS =  Entropy has the units Joules/Kelvin

26.  Named after physicist J. Willard Gibbs, free energy is the maximum amount of energy available during a chemical reaction.  Gibbs Free Energy Equation:  ΔG = ΔH – TΔS  When a reaction occurs at standard conditions (298 K and 1atm)  ΔG o = ΔH o - ΔS o

27.  ΔG = ΔH – TΔS  A reaction where ΔH is negative and ΔS is positive will always be spontaneous.  N 2 (g) + 3 H 2 (g)  2 NH 3 (g)  ΔH = -91.8 kJΔS = -197 J/k  ΔG =  ΔG = -33.1 kJ

28. ΔHΔSΔGReaction Spontaneity

29.  For a process ΔH is 145 kJ and ΔS is 322 J/K. Calculate ΔG for this reaction at 298 K. Is it spontaneous?